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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    Let a random variable X have a binomial distribution with mean 8 and variance 4. If \[P(x\le 2)=\frac{k}{{{2}^{16}}},\]then k is equal to:                 [JEE Main Held on 12-4-2019 Morning]

    A) 17                               

    B) 1

    C) 121                 

    D) 137

    Correct Answer: D

    Solution :

    \[np=8\] \[npq=4\] \[q=\frac{1}{2}\Rightarrow p=\frac{1}{2}\]            \[n=16\] \[p(x=r){{=}^{16}}{{C}_{r}}{{\left( \frac{1}{2} \right)}^{16}}\] \[p(x\le 2)=\frac{^{16}{{C}_{0}}{{+}^{16}}{{C}_{1}}{{+}^{16}}{{C}_{2}}}{{{2}^{16}}}\]\[=\frac{137}{{{2}^{16}}}\]


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